212 CHAPTER 5 Review Exercises Because ðx 0 , y 0 Þ is a solution of S 1 , a 1 x 0 + b 1 y 0 = c 1 , a 2 x 0 + b 2 y 0 = c 2 . Thus, ða 1 + ba 2 Þx 0 + ðb 1 + bb 2 Þy 0 = c 1 + bc 2 . So ðx 0 , y 0 Þ is a solution of S 2 . Assume now that ðx 0 , y 0 Þ is a solution of S 2 . Is it a solution of S 1 ? By definition of solution, ðx 0 , y 0 Þ satisfies both equations of the system S 2 . Thus, ðx 0 , y 0 Þ satisfies the first equation of S 1 as well. Therefore, we must prove that it satisfies the second equation of S 1 . By hypothesis ða 1 + ba 2 Þx 0 + ðb 1 + bb 2 Þy 0 = c 1 + bc 2 . We can rewrite the left-hand side of the second equation of S 2 as ða 1 + ba 2 Þx 0 + ðb 1 + bb 2 Þy 0 = ða 1 x 0 + b 1 y 0 Þ + bða 2 x 0 + b 2 y 0 Þ: Because ðx 0 , y 0 Þ is a solution of S 2 , the left-hand side of the equation is equal to c 1 + bc 2 , and the first expression in the parentheses in the right-hand side of the equation is equal to c 1 . Therefore, we obtain c 1 + bc 2 = c 1 + bða 2 x 0 + b 2 y 0 Þ. This equality implies that a 2 x 0 + b 2 y 0 = c 2 since b is a nonzero number. This proves that ðx 0 , y 0 Þ is a solution of S 1 . Use of Counterexamples 9. (Let us consider the statement. It seems to suggest that the "growth" of f should be cancelled by the "drop" of g. But the two functions could increase and decrease at different rates. Therefore, the statement does not seem to be true.) Let us construct a counterexample, using simple functions. We could try to use linear functions. Consider f ðxÞ = x + 1 and gðxÞ = -3x. Clearly, f is increasing and g is decreasing. (If you wish to do so, prove these claims.) Their sum is hðxÞ = -2x + 1, which is decreasing. 10.